Application

Of

Differential Equation in Our Real Life

Introduction:

A differential equation is a mathematical equation for an unknown function of

one or several variables that relates the values of the function itself and its derivatives

of various orders. Differential equations play a prominent role in engineering, physics,

economics, and other disciplines. One thing that will never change is the fact that the

world is constantly changing. Mathematically, rates of change are described by

derivatives. If we try and use math’s to describe the world around us like the growth

of plant, the growth of population, the fluctuations of the stock market, the spread of

diseases, or physical forces acting on an object. Most real life differential equation

needs to be solved numerically and many methods have been developed over the last

century and half and the goal has been to find methods that work for large classes of

differential equations. There seems to have been very little work published that

examines methods specialized to a single Differential Equation. By using Differential

Equation we can easily solve our day to day life problems.

Problem:

Influenza virus is one of the main problems in Bangladesh. Many people in our

country are affected by this virus. The person carrying an influenza virus returns to an

isolated village of 500 peoples. It is assumed that the rate at which the virus spreads is

proportional not only to the number of infected peoples but also to the people not

infected. Find the number of infected people after 5 months when it is further

observed that after 3 months. Number of infected peoples in 3 months is 30.

Mathematical Formulation:

Let Ni denote the number of infected people at any time t, N is the total number of

people and Nt is the time period.

Assuming that no one leaves the village throughout the duration of this disease, now

we can solve the initial value problem.

).........().........( iNNkNdt

dNii

i

The initial condition is, N (0) = 1

Conditions:

1. When t = 0 then Ni = 1

2. When t = 3 then Ni = 30

Solution:

Equation (i) is separable. Separating variables, we have

)..(....................)(

iikdtNNN

dN

ii

i

Integrating equation (ii),

dtkNNN

dN

ii

i

)(

dtkdN

NNNNi

ii

111

dtkNdN

NNNi

ii

11

ckNtNNN ii )ln(ln

ckNtNN

N

i

i

ln

kNt

i

i AeNN

N

kNti AeN

N 1

1 kNti AeN

N

).......(....................1

iiiAe

NN

kNti

When t=0 then Ni=1

From equation (iii), now we can write,

1

5001

A

5001 A

499 A

When t = 3 then Ni = 30 By using this condition now we can determine k from eq.(iii),

35004991

50030

ke

500)4991(30 1500 ke

03.01500 ke

00234.0 k

Then the equation (iii) becomes,

).........(....................1499

50017.1

ive

Nti

When t = 5 then the equation (iv) becomes,

517.14991

500

e

N i

or, 85.54991

500

e

N i

=205 peoples

Number of infected peoples during a time interval:

Period of time, Nt (Months) Number of infected people, Ni

3 30

4 88

5 205

6 345

7 439

8 479

9 493

10 497

Interpretation of Result:

The graph shows that there is a gradual increase in the number of infected

peoples. Day by day, most of the people of this village are infected this disease.

Nu

mb

er o

f In

fect

ed

Peo

ple

Time Period